Trigonometry - AS-Level Maths

Contributed by:

Sharp Tutor Mon, Jan 17, 2022 08:50 PM UTC

SECTIONS:
1. The sine, cosine, and tangent of any angle
2. The graphs of sin θ, cos θ, and tan θ
3. Exact values of trigonometric functions
4. Trigonometric equations
5. Trigonometric identities
6. Examination-style questions

1. AS-Level Maths:
Core 2
for Edexcel
C2.4 Trigonometry 1
These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page.
This icon indicates the slide contains activities created in Flash. These activities are not editable.
For more detailed instructions, see the Getting Started presentation.
1 of 47 © Boardworks Ltd 2005

2. The sine, cosine and tangent of any angle
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θ and tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
2 of 47 © Boardworks Ltd 2005

3. The three trigonometric ratios
The three trigonometric ratios, sine, cosine and tangent, can
be defined using the ratios of the sides of a right-angled
triangle as follows:
Opposite
Sin θ =
Hypotenuse SOH
H
O Y
P P
P O Adjacent
T
E
N
Cos θ =
Hypotenuse CAH
I U
S
T E
E Opposite
θ Tan θ =
Adjacent TOA
ADJACENT
Remember: S O H C A H T O A
3 of 47 © Boardworks Ltd 2005

4. The sine, cosine and tangent of any angle
These definitions are limited because a right-angled triangle
cannot contain any angles greater than 90°.
To extend the three trigonometric ratios to include angles
greater than 90° and less than 0° we consider the rotation of a
straight line OP of fixed length r about the origin O of a
coordinate grid.
y Angles are then measured
P(x, y)
anticlockwise from the
r θ positive x-axis.
α
O x For any angle θ there is an
associated acute angle α
between the line OP and
the x-axis.
4 of 47 © Boardworks Ltd 2005

5. The sine, cosine and tangent of any angle
The three trigonometric ratios are then given by:
y
sin =
r
x
cos =
r
y
tan =
x
The x and y coordinates can be positive or negative, while r is
always positive.
This means that the sign of the required ratio will depend on
the sign of the x-coordinate and the y-coordinate of the point P.
5 of 47 © Boardworks Ltd 2005

6. The sine, cosine and tangent of any angle
If we take r to be 1 unit long then these ratios can be written
y
sin = = y
1
x
cos = = x
1
y sin
tan =  tan =
x cos 
The relationship between θ measured from the positive x-axis
and the associated acute angle α depends on the quadrant
that θ falls into.
For example, if θ is between 90° and 180° it will fall into the
second quadrant and α will be equal to (180 – θ)°.
6 of 47 © Boardworks Ltd 2005

7. The sine of any angle
If the point P is taken to revolve about a unit circle then sin θ is
given by the y-coordinate of the point P.
7 of 47 © Boardworks Ltd 2005

8. The cosine of any angle
If the point P is taken to revolve about a unit circle then cos θ
is given by the x-coordinate of the point P.
8 of 47 © Boardworks Ltd 2005

9. The tangent of any angle
Tan θ is given by the y-coordinate of the point P divided by the
9 of 47 © Boardworks Ltd 2005

10. The tangent of any angle
Tan θ can also be given by the length of tangent from the point
P to the x-axis.
10 of 47 © Boardworks Ltd 2005

11. Remember CAST
We can use CAST to remember in which quadrant each of
the three ratios are positive.
2nd quadrant 1st quadrant
S
Sine is positive A
All are positive
3rd quadrant 4th quadrant
T
Tangent is positive C
Cosine is positive
11 of 47 © Boardworks Ltd 2005

12. The sine, cosine and tangent of any angle
The sin, cos and tan of angles in the first quadrant are positive.
In the second quadrant: sin θ = sin α
cos θ = –cos α
tan θ = –tan α
In the third quadrant: sin θ = –sin α where α is the
cos θ = –cos α associated
tan θ = tan α acute angle.
In the fourth quadrant: sin θ = –sin α
cos θ = cos α
tan θ = –tan α
12 of 47 © Boardworks Ltd 2005

13. The sine, cosine and tangent of any angle
The value of the associated acute angle α can be found using
a sketch of the four quadrants.
For angles between 0° and 360° it is worth remembering that:
when 0° < θ < 90°, α = θ
when 90° < θ < 180°, α = 180° – θ
when 180° < θ < 270°, α = θ – 180°
when 270° < θ < 360°, α = 360° – θ
For example, if θ = 230° we have:
α = 230° – 180° = 50°
230° is in the third quadrant where only tan is positive and so:
sin 230° = –sin 50°
cos 230° = –cos 50°
tan 230° = tan 50°
13 of 47 © Boardworks Ltd 2005

14. The graphs of sin θ, cos θ and tan θ
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θ and tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
14 of 47 © Boardworks Ltd 2005

15. The graph of y = sin θ
15 of 47 © Boardworks Ltd 2005

16. The graph of y = sin θ
The graph of sine is said to be periodic since it repeats itself
every 360°.
We can say that the period of the graph y = sin θ is 360°.
Other important features of the graph y = sin θ include the
fact that:
It passes through the origin, since sin 0° = 0.
The maximum value is 1 and the minimum value is –1.
Therefore, the amplitude of y = sin θ is 1.
It has rotational symmetry about the origin. In other words, it
is an odd function and so sin (–θ) = –sin θ.
16 of 47 © Boardworks Ltd 2005

17. The graph of y = cos θ
17 of 47 © Boardworks Ltd 2005

18. The graph of y = cos θ
Like the graph of y = sin θ the graph of y = cos θ is periodic
since it repeats itself every 360°.
We can say that the period of the graph y = cos θ is 360°.
Other important features of the graph y = cos θ include the
fact that:
It passes through the point (0, 1), since cos 0° = 1.
The maximum value is 1 and the minimum value is –1.
Therefore, the amplitude of y = cos θ is 1.
It is symmetrical about the y-axis. In other words, it is an
even function and so cos (–θ) = cos θ.
It the same as the graph of y = sin θ translated left 90°. In
other words, cos θ = sin (90° – θ).
18 of 47 © Boardworks Ltd 2005

19. The graph of y = tan θ
19 of 47 © Boardworks Ltd 2005

20. The graph of y = tan θ
You have seen that the graph of y = tan θ has a different
shape to the graphs of y = sin θ and y = cos θ.
Important features of the graph y = tan θ include the fact that:
It is periodic with a period of 180°.
It passes through the point (0, 0), since tan 0° = 0.
Its amplitude is not defined, since it ranges from +∞ to –∞.
It is symmetrical about the origin. In other words, it is an odd
function and so tan (–θ) = –tan θ.
tan θ is not defined for θ = ±90°, ±270°, ±450°, … that is, for
odd multiples of 90°. The graph y = tan θ therefore contains
asymptotes at these points.
20 of 47 © Boardworks Ltd 2005

21. Exact values of trigonometric functions
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θ and tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
21 of 47 © Boardworks Ltd 2005

22. Sin, cos and tan of 45°
A right-angled isosceles triangle has two acute angels of 45°.
Suppose the equal sides are of
45° unit length.
2 1 Using Pythagoras’ theorem:
2 2
The hypotenuse  1 1
45°
 2
1
We can use this triangle to write exact values for sin, cos and
tan 45°:
1 1
sin 45° = cos 45° = tan 45° = 1
2 2
22 of 47 © Boardworks Ltd 2005

23. Sin, cos and tan of 30°
Suppose we have an equilateral triangle of side length 2.
If we cut the triangle in half then we have
60° a right-angled triangle with acute angles
30° of 30° and 60°.
2 2
3 Using Pythagoras’ theorem:
2 2
60° 60° The height of the triangle  2  1
1 2  3
We can use this triangle to write exact values for sin, cos and
tan 30°:
1 3 1
sin 30° = cos 30° = tan 30° =
2 2 3
23 of 47 © Boardworks Ltd 2005

24. Sin, cos and tan of 60°
Suppose we have an equilateral triangle of side length 2.
If we cut the triangle in half then we have
60° a right-angled triangle with acute angles
30° of 30° and 60°.
2 2
3 Using Pythagoras’ theorem:
2 2
60° 60° The height of the triangle  2  1
1 2  3
We can also use this triangle to write exact values for sin, cos
and tan 60°:
3 1
sin 60° = cos 60° = tan 60° = 3
2 2
24 of 47 © Boardworks Ltd 2005

25. Sin, cos and tan of 30°, 45° and 60°
The exact values of the sine, cosine and tangent of 30°, 45°
and 60° can be summarized as follows:
30° 45° 60°
1 1 3
sin
2 2 2
3 1 1
cos 2 2
2
1
tan 1 3
3
Use this table to write the exact value of cos 135°
1
cos 135° = –cos 45° = 2
25 of 47 © Boardworks Ltd 2005

26. Sin, cos and tan of 30°, 45° and 60°
Write the following ratios exactly:
1
1) cos 300° = 2
2) tan 315° = -1
1
3) tan 240° = 3 4) sin –330° = 2
3
5) cos –30° = 6) tan –135° = 1
2
1 1
7) sin 210° = 2
8) cos 315° = 2
26 of 47 © Boardworks Ltd 2005

27. Trigonometric equations
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θ and tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
27 of 47 © Boardworks Ltd 2005

28. Equations of the form sin θ = k
Equations of the form sin θ = k, where –1 ≤ k ≤ 1, have an
infinite number of solutions.
If we use a calculator to find arcsin k (or sin–1 k) the calculator
will give a value for θ between –90° and 90°.
There is one and only one solution is this range.
This is called the principal solution of sin θ = k.
Other solutions in a given range can be found using the
graph of y = sin θ or by considering the unit circle.
For example:
Solve sin θ = 0.7 for –360° < θ < 360°
arcsin 0.7 = 44.4° (to 1 d.p.)
28 of 47 © Boardworks Ltd 2005

29. Equations of the form sin θ = k
Using the graph of y = sin θ between –360° and 360° and the
line y = 0.7 we can locate the other solutions in this range.
y = sin θ
y = 0.7
–315.6° –224.4° 44.4° 135.6°
So the solutions to sin θ = 0.7 for –360° < θ < 360° are:
θ = –315.6°, –224.4°, 44.4°, 135.6° (to 1 d.p)
29 of 47 © Boardworks Ltd 2005

30. Equations of the form sin θ = k
30 of 47 © Boardworks Ltd 2005

31. Equations of the form sin θ = k
We could also solve sin θ = 0.7 for –360° < θ < 360° by
considering angles in the first and second quadrants of a unit
circle where the sine ratio is positive.
Start by sketching the principal solution 44.4° in the first
Next, sketch the associated acute
–224.4° –315.6° angle in the second quadrant.
135.6° 44.4° Moving anticlockwise from the
x-axis gives the second solution:
180° – 44.4° = 135.6°
Moving clockwise from the
x-axis gives the third and fourth
solutions:
–(180° + 44.4°) = –224.4°
–(360° – 44.4°) = –315.6°
31 of 47 © Boardworks Ltd 2005

32. Equations of the form cos θ = k and tan θ = k
Equations of the form cos θ = k, where –1 ≤ k ≤ 1, and
tan θ = k, where k is any real number, also have infinitely
many solutions. For example:
Solve tan θ = –1.5 for –360° < θ < 360°
Using a calculator, the principal solution is θ = –56.3° (to 1 d.p.)
123.7° Now look at angles in the second and
–236.3°
fourth quadrants of a unit circle where
the tangent ratio is negative.
This gives us four solutions in the
–56.3°
range –360° < θ < 360°:
303.7°
θ = –236.3°, –56.3°, 123.7°, 303.7°
32 of 47 © Boardworks Ltd 2005

35. Equations involving multiple or compound angles
Multiples angles are angles of the form aθ where a is a given
Compound angles are angles of the form (θ + b) where b is a
given constant.
When solving trigonometric equations involving these types of
angles, care should be taken to avoid ‘losing’ solutions.
Solve cos 2θ = 0.4 for –180° < θ < 180°
Start by changing the range to match the multiple angle:
–180° < θ < 180°
–360° < 2θ < 360°
Next, let x = 2θ and solve the equation cos x = 0.4 in the range
–360° ≤ x ≤ 360°.
35 of 47 © Boardworks Ltd 2005

36. Equations involving multiple or compound angles
Now, using a calculator: x = 66.4° (to 1 d.p.)
–293.6° Using the unit circle to find the values
66.4° of x in the range –360° ≤ x ≤ 360°
gives:
x = 66.4°, 293.6°, –66.4°, –293.6°
But x = 2θ so:
293.6°
–66.4° θ = 33.2°, 146.8°, –33.2°, –146.8°
This is the complete solution set in the range –180° < θ < 180°.
36 of 47 © Boardworks Ltd 2005

37. Equations involving multiple or compound angles
Solve tan(θ + 25°)= 0.8 for 0° < θ < 360°
Start by changing the range to match the compound angle:
0° < θ < 360°
25° < θ + 25° < 385°
Next, let x = θ + 25° and solve the equation tan x = 0.8 in the
range 25° < x < 385°.
38.7°
Using a calculator: x = 38.7° (to 1 d.p.)
Using the unit circle to find the values
of x in the range 25° ≤ x ≤ 385° gives:
x = 38.7°, 218.7° (to 1 d.p.)
218.7°
But x = θ + 25° so:
θ = 13.7°, 193.7.7° (to 1 d.p.)
37 of 47 © Boardworks Ltd 2005

38. Trigonometric equations involving powers
Sometimes trigonometric equations involve powers of sin θ,
cos θ and tan θ. For example:
Solve 4sin2θ – 1= 0 for –180° ≤ θ ≤ 180°
Notice that (sin θ)2 is usually written as sin2θ.
4 sin2   1= 0
4 sin2  = 1
sin2  = 41
sin = 21
When sin θ = 21 , θ = 30°, 150°
When sin θ = – 21 , θ = –30°, –150°
So the full solution set is θ = –150°, –30°, 30°, 150°
38 of 47 © Boardworks Ltd 2005

39. Trigonometric equations involving powers
Solve 3cos2θ – cos θ = 2 for 0° ≤ θ ≤ 360°
Treat this as a quadratic equation in cos θ.
3cos2θ – cos θ = 2
3cos2θ – cos θ – 2 = 0
(cos θ – 1)(3cos θ + 2) = 0
cos θ = 1 or cos θ = – 32
When cos θ = 1, θ = 0°, 360°
When cos θ = – 32 , θ = 131.8°, 22.8.2°
So the full solution set is θ = 0°, 131.8°, 228.2°, 360°
39 of 47 © Boardworks Ltd 2005

40. Trigonometric identities
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θ and tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
40 of 47 © Boardworks Ltd 2005

41. Trigonometric identities
Two important identities that must be learnt are:
sin
 tan (cos 0)
cos
sin2θ + cos2θ ≡ 1
The symbol ≡ means “is identically equal to” although an
equals sign can also be used.
An identity, unlike an equation, is true for every value of the
given variable so, for example:
sin24° + cos24° ≡ 1, sin267° + cos267° ≡ 1, sin2π + cos2π ≡ 1, etc.
41 of 47 © Boardworks Ltd 2005

42. Trigonometric identities
We can prove these identities by considering a right-angled
y x
sin = and cos =
r r r
y sin y
y
 = r = = tan as required.
θ cos x r x
x 2 2
y
   x
Also: sin2  + cos2  =   +  
r r
x2 + y 2
=
r2
But by Pythagoras’ theorem x2 + y2 = r2 so:
2
r
sin2  + cos2  = 2 = 1 as required.
r
42 of 47 © Boardworks Ltd 2005

43. Trigonometric identities
One use of these identities is to simplify trigonometric
equations. For example:
Solve sin θ = 3 cos θ for 0° ≤ θ ≤ 360°
sin 3cos
Dividing through by cos θ: =
cos cos
tan = 3
Using a calculator, the principal solution is θ = 71.6° (to 1 d.p.)
71.6° So the solutions in the given range are:
θ = 71.6°, 251.6° (to 1 d.p.)
251.6°
43 of 47 © Boardworks Ltd 2005

44. Trigonometric identities
Solve 2cos2θ – sin θ = 1 for 0 ≤ θ ≤ 360°
We can use the identity cos2θ + sin2 θ = 1 to rewrite this
equation in terms of sin θ.
2(1 – sin2 θ) – sin θ = 1
2 – 2sin2θ – sin θ = 1
2sin2θ + sin θ – 1 = 0
(2sin θ – 1)(sin θ + 1) = 0
So: sin θ = 0.5 or sin θ = –1
If sin θ = 0.5, θ = 30°, 150°
If sin θ = –1, θ = 270°
44 of 47 © Boardworks Ltd 2005

45. Examination-style questions
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θ and tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
45 of 47 © Boardworks Ltd 2005

46. Examination-style question
Solve the equation
3 sin θ + tan θ = 0
for θ for 0° ≤ θ ≤ 360°
sin
Rewriting the equation using tan  :
cos
sin
3 sin + =0
cos
3 sin cos + sin = 0
sin (3cos +1) = 0
So sin θ = 0 or 3 cos θ + 1 = 0
3 cos θ = –1
cos θ = – 31
46 of 47 © Boardworks Ltd 2005

47. Examination-style question
In the range 0° ≤ θ ≤ 360°,
when sin θ = 0, θ = 0°, 180°, 360°.
1
when cos θ = – 3 :
cos–1 – 31 = 109.5° (to 1 d.p.)
cos θ is negative in the 2nd and 3rd quadrants so the second
solution in the range is:
109.5°
θ = 250 .5° (to 1 d.p.)
So the complete solution set is:
250.5° θ = 0°, 180°, 360°, 109.5°, 250 .5°
47 of 47 © Boardworks Ltd 2005

Book a Trial Class

Download